package com.hit.basmath.learn.binary_search_tree;

import com.hit.common.TreeNode;

/**
 * 450. Delete Node in a BST
 * <p>
 * Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.
 * <p>
 * Basically, the deletion can be divided into two stages:
 * <p>
 * 1. Search for a node to remove.
 * 2. If the node is found, delete the node.
 * <p>
 * Note: Time complexity should be O(height of tree).
 * <p>
 * Example:
 * <p>
 * root = [5,3,6,2,4,null,7]
 * <p>
 * key = 3
 * <p>
 * 5
 * / \
 * 3   6
 * / \   \
 * 2   4   7
 * <p>
 * Given key to delete is 3. So we find the node with value 3 and delete it.
 * <p>
 * One valid answer is [5,4,6,2,null,null,7], shown in the following BST.
 * <p>
 * 5
 * / \
 * 4   6
 * /     \
 * 2       7
 * <p>
 * Another valid answer is [5,2,6,null,4,null,7].
 * <p>
 * 5
 * / \
 * 2   6
 * \   \
 * 4   7
 */
public class _450 {

    /**
     * Recursion method
     *
     * @param root pending root node
     * @param key  target value
     * @return tree node
     */
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) {
            return null;
        }
        if (key < root.val) {
            root.left = deleteNode(root.left, key);
        } else if (key > root.val) {
            root.right = deleteNode(root.right, key);
        } else {
            if (root.left == null) {
                return root.right;
            } else if (root.right == null) {
                return root.left;
            }

            TreeNode minNode = findMin(root.right);
            root.val = minNode.val;
            root.right = deleteNode(root.right, root.val);
        }
        return root;
    }

    private TreeNode findMin(TreeNode node) {
        while (node.left != null) {
            node = node.left;
        }
        return node;
    }
}
